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Single Pole IIR Filter Frequency Response
July 6, 2022

Table of Contents

Introduction

This blog post demonstrates how to find the frequency response of the single pole IIR filter and then plots the magnitude and phase responses for multiple \alpha values.

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Single Pole IIR Impulse Response

The single pole infinite impulse response (IIR) filter is an incredibly efficient filter!

The impulse response of the single pole IIR is

(1)   \begin{equation*}y[n] = \alpha x[n] + (1-\alpha)y[n-1]\end{equation*}

The filter uses two weights, \alpha and 1-\alpha. The weight \alpha is applied to the input signal x[n] and the value is typically small ( \alpha < 0.1 ). The weight 1-\alpha is much larger than \alpha and is applied to the feedback y[n-1]. The smaller the \alpha the more impact the feedback has on the output and the smoother (and more narrowband) the filter.

The single pole IIR may also be referred to as an exponential moving average (EMA) filter.

Single Pole IIR Frequency Response

The Z-transform is applied to (1),

(2)   \begin{equation*}Y(z) = \alpha X(z) + (1-\alpha) Y(z)z^{-1}.\end{equation*}

Equation (2) is simplified by gathering like terms,

(3)   \begin{equation*}Y(z) - (1-\alpha)Y(z)z^{-1} = \alpha X(z),\end{equation*}

(4)   \begin{equation*}Y(z)(1 - (1-\alpha)z^{-1}) = \alpha X(z).\end{equation*}

The transfer function for (1) is written as the ratio of Y(z) to X(z),

(5)   \begin{equation*}H(z) = \frac{Y(z)}{X(z)} = \frac{\alpha}{1 - (1-\alpha)z^{-1}}\end{equation*}

The frequency response H\left( e^{j\omega}\right) is created by substituting

(6)   \begin{equation*}z=e^{j\omega}\end{equation*}

into (5) resulting in

(7)   \begin{equation*}H\left(e^{j\omega}\right) = \frac{\alpha}{1 - (1-\alpha)e^{-j\omega}}.\end{equation*}

Simplify (7) by definining

(8)   \begin{equation*}\beta = 1-\alpha\end{equation*}

such that the frequency response is now

(9)   \begin{equation*}H\left(e^{j\omega}\right) = \frac{\alpha}{1-\beta e^{-j\omega}}.\end{equation*}

Magnitude of the Frequency Response

The magnitude of the frequency response (9) is

(10)   \begin{equation*}\left| H\left( e^{j\omega} \right) \right| = \sqrt{ H\left(e^{j\omega})H^*\left(e^{j\omega}\right)}.\end{equation*}

Substituting (9) into (10),

(11)   \begin{equation*}\left|H\left(e^{j\omega}\right)\right| = \frac{\alpha}{\sqrt{\left(1 - \beta e^{-j\omega}\right)\left(1 - \beta e^{j\omega}\right)}}.\end{equation*}

Example Plots

The magnitude and phase of the frequency response (9) is given below. The plots show that as \alpha decreases the filter becomes more narrowband and has more phase linearity.

A smaller alpha value leads to a more narrow filter bandwidth for the single pole IIR.
A smaller alpha value leads to a more narrow filter bandwidth for the single pole IIR.
A smaller alpha value leads to more phase linearity for the single pole IIR filter.
A smaller alpha value leads to more phase linearity for the single pole IIR filter.

Conclusion

This blog demonstrated how to find the frequency response of the a single pole IIR filter and plotted the phase and magnitude responses for several \alpha values.

Check out these other blogs on DSP:

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